Question 1039138
.
8 sin^2 x + 3 sin 2x = 4
Solve for all angles btw 0 and 360 which satisfy the equation. 
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
{{{8sin^2(x) + 3sin(2x)}}} = {{{4}}},  --->

{{{(8sin^2(x) -4) + 3sin(2x)}}} = {{{0}}},  --->

{{{4(2sin^2(x) -1) + 3sin(2x)}}} = {{{0}}},  --->    (replace 1 by {{{sin^2(x) + cos^2(x)}}})

{{{4(2sin^2(x) -(sin^2(x) + cos^2(x))) + 3sin(2x)}}} = {{{0}}},  --->

{{{4(sin^2(x) - cos^2(x)) + 3sin(2x)}}} = {{{0}}},  --->  (replace {{{sin^2(x) - cos^2(x)}}}  by  -cos(2x)) 

-4cos(2x) + 3sin(2x) = 0  --->  4cos(2x) = 3sin(2x)  --->  tan(2x) = {{{4/3}}}  --->  
2x = {{{arctan(4/3)}}}  and/or  2x = {{{arctan(4/3) + pi}}}  --->  

x = {{{(1/2)*arctan(4/3)}}}  and/or  x = {{{(1/2)*arctan(4/3) +pi/2}}}.
</pre>

Now use your calculator to convert arctan into degrees.


Solved.